Answer:
See attachment for graph.
The function is not continuous.
Step-by-step explanation:
Piecewise functions have multiple pieces of curves/lines where each piece corresponds to its definition over an interval.
Given piecewise function:
[tex]f(x)=\begin{cases} x-3 \;\;\;\;\: \text{if}\;\;\;x \leq -2\\4x+5 \;\;\; \text{if}\;\; \;x > -2 \end{cases}[/tex]
Therefore, the function has two definitions:
- f(x) = x - 3 when x is less than or equal to -2.
- f(x) = 4x + 5 when x is greater than -2.
When graphing piecewise functions:
- Use an open circle where the value of x is not included in the interval.
- Use a closed circle where the value of x is included in the interval.
- Use an arrow to show that the function continues indefinitely.
First piece of the function
Substitute x = -2 into the first function:
[tex]\implies f(-2)=-2-3=-5[/tex]
As the interval for the first equation is x ≤ -2, it includes the value of x = -2. Therefore, place a closed circle at point (-2, -5).
To help graph the line, find another point on the line by inputting another value of x that is less than -2 into the same function:
[tex]\implies f(-5)=-5-3=-8[/tex]
Plot point (-5, -8) and draw a straight line from the closed circle at (-2, -5) through (-5, -8). Add an arrow at the other endpoint to show it continues indefinitely as x → -∞.
Second piece of the function
Substitute x = -2 into the second function:
[tex]\implies f(-2)=4(-2)+5=-3[/tex]
As the interval for the second equation is x > -2, it does not include the value of x = -2. Therefore, place an open circle at point (-2, -3).
To help graph the line, find another point on the line by inputting another value of x that is more than -2 into the same function:
[tex]\implies f(1)=4(1)+5=9[/tex]
Plot point (1, 9) and draw a straight line from the open circle at (-2, -3) through (1, 9). Add an arrow at the other endpoint to show it continues indefinitely as x → ∞.
See attachment for the graph.
[tex]\boxed{\begin{minipage}{8cm}\underline{Determining if a function is continuous at $x=a$}\\\\Condition 1: $f(a)$ exists\\\\Condition 2: $\displaystyle \lim_{x \to a}f(x)$ exists at $x=a$\\\\Condition 3: $\displaystyle \lim_{x \to a}f(x)=f(a)$\\\end{minipage}}[/tex]
If all three conditions are satisfied, the function is continuous at x = a.
If any one of the conditions is not satisfied, the function is not continuous at x = a.
To determine whether or not the given piecewise function is continuous, find if the function is continuous at x = -2.
Condition 1
Does f(-2) exist? Yes → f(-2) = -5
Condition 2
[tex]\textsf{Does}\;\;\displaystyle \lim_{x \to -2} f(x)\;\; \sf exist\;at\;\;x=-2?[/tex]
To the left of x =- 2, f(x) = x - 3
To the right of x = -2 , f(x) = 4x + 5
Evaluate the left and right limits as x approaches -2:
[tex]\displaystyle \lim_{x \to -2^-}f(x)=\lim_{x \to -2^-} -2-3=-5[/tex]
[tex]\displaystyle \lim_{x \to -2^+}f(x)=\lim_{x \to -2^+} 4(-2)+5=-3[/tex]
[tex]\textsf{As}\;\;\displaystyle \lim_{x \to -2^-} f(x) \neq \lim_{x \to -2^+} f(x), \;\; \lim_{x \to -2} f(x)\;\; \textsf{does not exist}.[/tex]
As condition 2 fails, there is no need to proceed to condition 3.
Therefore, the function is not continuous.
